Question

exercises 2.3 the limit laws
score: 18.75/21 answered: 19/21
question 20
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find the limit.
\\(\lim_{x\to0}\frac{\tan(7x)}{x}=\\)

Explanation:

Step1: Recall tangent identity

We know that $\tan(7x)=\frac{\sin(7x)}{\cos(7x)}$, so the limit becomes $\lim_{x
ightarrow0}\frac{\sin(7x)}{x\cos(7x)}$.

Step2: Rewrite the limit

$\lim_{x
ightarrow0}\frac{\sin(7x)}{x\cos(7x)}=\lim_{x
ightarrow0}\frac{\sin(7x)}{x}\cdot\lim_{x
ightarrow0}\frac{1}{\cos(7x)}$.

Step3: Use the limit - rule $\lim_{u

ightarrow0}\frac{\sin u}{u} = 1$
Let $u = 7x$. As $x
ightarrow0$, $u
ightarrow0$. And $\lim_{x
ightarrow0}\frac{\sin(7x)}{x}=7\lim_{x
ightarrow0}\frac{\sin(7x)}{7x}=7\times1 = 7$.

Step4: Evaluate $\lim_{x

ightarrow0}\frac{1}{\cos(7x)}$
Since $\cos(0)=1$, $\lim_{x
ightarrow0}\frac{1}{\cos(7x)}=\frac{1}{\cos(0)} = 1$.

Step5: Calculate the original limit

$\lim_{x
ightarrow0}\frac{\sin(7x)}{x}\cdot\lim_{x
ightarrow0}\frac{1}{\cos(7x)}=7\times1=7$.

Answer:

$7$